Quick question about integration

  • Thread starter Thread starter dand5
  • Start date Start date
  • Tags Tags
    Integration
dand5
Messages
28
Reaction score
0
I have a quick question about integration after a change of variables has been made.

Suppose there is a function R(t_{1},t_{2}) that actually just
depends on the difference t_{1} - t_{2}. The goal is then to
simplify the following integral:

<br /> \frac{1}{T^{2}}\int^{T}_{0}\int^{T}_{0} R(t_{1},t_{2}) dt_{1}dt_{2}<br />

by using the substitution t_{1}&#039; = t_{1} and t_{2}&#039;= t_{1} - \tau.

A straight substitution yields:
<br /> \frac{1}{T^{2}}\int \int^{T}_{0} R(\tau) dt_{1}&#039;(dt_{1}&#039; - d\tau)<br />

I am uncertain about two things:

1) the integration bounds on the outer integral after the substitution has been made
2) whether or not dt_{1}&#039; in the outer integral is zero since
dt_{1} is held constant when integrating over dt_{2} before the substitution was made.

As a heads up the final result is supposed to be:
<br /> \frac{1}{T^{2}}\int^{T}_{-T}\left(T-\left|\tau\right|\right)R(\tau) d\tau<br />

Thanks in advance for any responses.
 
Physics news on Phys.org
To simplify writing I will use s instead of tau.

Let s=t1-t2, t=t2. Then the differential looks like R(s)dsdt, with s limits -t,T-t and the t limits 0,T.

Next reverse the order of integration. The t limit is split into 2 parts. For s<0, the limits are -s and T, while for s>0, the limits are 0 and T-s. The s limits are -T and T.

When you integrate with respect to t, you get T-|s|.
 
ok I see it now, Thanks. Also, the determinant of the Jacobian of the coordinate transformation is 1, so I guess it is correct to say that dt1dt2=dsdt.
 
Back
Top